Binary Phase Diagram (Txy and xy)
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- Опубликовано: 12 окт 2024
- Organized by textbook: learncheme.com/
Explains how to generate phase diagrams and read them for use in separation calculations. Made by faculty at the University of Colorado Boulder, Department of Chemical & Biological Engineering. Check out our separations/mass transfer playlist: • Separations/Mass Transfer
![BOLIVIA vs. COLOMBIA [1-0] | RESUMEN | ELIMINATORIAS SUDAMERICANAS | FECHA 9](http://i.ytimg.com/vi/zg15uVWjuf4/mqdefault.jpg)
It's amazing how much information is packed into one plot. Beautiful, really.
I have a question from above graph...
At 65℃ , how can the mole fraction of water in liquid is zero???
Plz answer
1. This is because the water is at 30.4 kPa where the boiling point of water is about 70 C. So in case of pure water(right bottom corner of graph) the boiling will start at 70 C. Instead of standard conditions of 101kPa where the boiling point is 100 C. 2. If you are looking at the left bottom corner of the graph it is for pure ethylene alcohol hence the mole fraction of water will be zero no matter how much you increase the temperature. I hope this answers your question
I can understand that there's no water at 163 C as it might have boiled/evaporated completely. However, I can't understand why there's no Ethylene Glycol at 68 C in both liquid and water phase? Isn't it the less volatile component?
See the table at 0:54. Different mixture compositions, have different bubble points and dew points. At 68 °C, what you are basically seeing is the bubble point being equal to the dew point because there is no mixture. It's just pure water. There is no ethylene glycol at the right-most point
What I do not understand conceptually is how you only have one phase as you heat the system ? At 3:02 you have 0.5 mol frac water, 0.5 mol frac ethylene glycol and as you heat the system you only have the liquid phase present? If I have a mole fraction of 0.5 water and 0.5 ethylene glycol in a sealed jar and I heat that system, won't I be pressurizing the vessel as more vapor is entering the headspace of the vessel as I heat from 65 to 85C and won't the mol fraction in the liquid phase be constantly changing? Can you tell me where I am going wrong here?
Thank you for the question. The answer is that before you get to approx. 86 degC, there is no vapor. There is only a liquid phase until you get to the bubble point when some vapor starts to form. Therefore, the mole fraction of the liquid is not changing before then. Within the 2-phase envelope, the mole fractions of both the liquid and vapor will change as the temperature increases. Does that answer your question?
@@LearnChemE I understand the two phase envelope. But don't all components have some vapor pressure as a function of temperature? The bubble point occurs only when the vapor pressure equals the external pressure, correct? Furthermore, In the example I gave, typically we are taught in a chemistry lab to not heat vessels that are closed (i.e. a flask with a stopper) because there is pressure buildup as the vapor pressure of water increases with temperature,which means there must be a vapor phase. Why is that not applicable here? Thanks for the response
@@bharathk9594I appreciate your continued interest in this question and please keep asking if I don't explain it well here. Before we hit 86 degC, there is no vapor, so there is no vapor pressure. However, in this case we have no air above the liquid. If we did, it would absolutely mean there is vapor and vapor pressure - and that would be the case in your labs. For safety reasons, do not heat a closed vessel. That is a different situation than the screencast above.
@@LearnChemE Thanks for the explanations. I suppose I did not quite understand how this vle data is used to model real-world scenarios, if there was no air above the liquid. Is this only applied in ideal scenarios ?
@ 4:23 At the first point on the dew point line, you have 0.5 liquid H2O and 0.97 vapor H2O so the balance is 0.5L+0.97V=0.5, and if we do the same balance for ethyl glycol we have 0.5L+0.03V=0.5 which doesn't make sense since the equations cannot cancel out.
You add liquid mole fraction to liquid mole fraction and vapour mole fraction to vapour mole fraction. You don't know the total number of moles in either phase so you can't just add them. They are fractions of a total (in their phase).
What did you use to plot this graph?
It's in Excel
Ok this sort of still does not make sense to me. If the amount of water in liquid phase decreases, then it is going to increase in the vapor phase, and so do behave its liquid and vapor molfractions. Where am I going wrong in here please ?
Good question. This is tricky. When the temperature increases, so does the amount of vapor. The mole fractions of water are decreasing in both phases, but the total moles of vapor and liquid are also changing, so the total amounts of water and ethylene glycol also remain the same. Check the second part of this video (starting around 3:30):
ruclips.net/video/-XcTEknC9Aw/видео.html
Enlightening
how do you determine which is the more volatile component.
ruclips.net/video/-XcTEknC9Aw/видео.html
the one that is lighter
The compound having higher vapor pressure and lower boiling point is highly volatile..
On the basis of vapour pressure,inside the liq if molecules more attracted to each other,they have less vapour pressure,more different to boil,therefore less volatile
@@jay.jay. It's more complicated than that. Both molecular mass is a factor, but so are intermolecular forces (H-bonds, van der waals etc). The correct answer is which ever component has the highest vapor pressure.